(In reply to
please read by Charlie)
I'm in a little hurry right now so this explanation might not be perfect. B doesn't have to encode the values of the 4 cards in to the card he picks. e.g. if the first audience member picks a, the second one b and so on, there are only 4!=24 possible ways to form an inequality group of these four numbers. a<b<c<d, b<a<c<d and so on.. B choses a card which just indicates which inequality these 4 cards posses and A is able to figure which card belongs to which member. This is something that was already mentioned in previous posts. The real difficulty of this problem (in my opinion) is to figure the way for B to choose a card that can be recognized by A and isn't already taken.
Although I would like to remind that in the original problem you only had to show this is possible not the way to do it. However it's much more pleasing to find the way rather than prove it can be done.

Posted by atheron
on 20070512 11:47:58 